mysql_fetch_array()的那些事
调试程序时出现这个问题：mysql_fetch_array() expects parameter 1 to be resource, boolean given in
请问是什么问题？
程序如下：
$result2=mysql_query(“select sum(score) as sum from ask_score where qid=”.$qid.”and answerid=”.$authid);
$author3 = mysql_fetch_array ( $result2, MYSQL_ASSOC );
$scoresum=$author3[“sum”];
$result3=mysql_query(“select count(*) as count from ask_score where qid=”.$qid);
$author4 = mysql_fetch_array ( $result3, MYSQL_ASSOC );
$scorenum=$author4[“count”];
ask_score表：
CREATE TABLE ask_socre(
qid int(10) unsigned NOT NULL default ‘0’,
authorid mediumint(8) unsigned NOT NULL default ‘0’,
answerid mediumint(8) unsigned NOT NULL default ‘0’,
vote_id mediumint(8) unsigned NOT NULL default ‘0’,
score int(2) uns
搞代gaodaima码igned NOT NULL default ‘0’,
PRIMARY KEY (qid,authorid)
);
——解决方案——————–
你的mysql_query返回了一个false，sql语句错误
改成mysql_query(…) or die(mysql_error()); 看sql语句有什么错误
——解决方案——————–
1.表名
2.加点空格，最好加上单引号
where qid=”.$qid.”and answerid=”.$
where qid=”.$qid.” and answerid=”.$
- PHP code
<?php$qid = '1';$authid = '1';$link = mysql_connect('127.0.0.1', 'root', '');mysql_select_db("test");$sql2= " select sum(score) as sum from ask_socre where qid='".$qid."' and answerid='".$authid."' ";$result2=mysql_query($sql2);echo $sql2;if($author3 = mysql_fetch_array ( $result2, MYSQL_ASSOC )){ var_dump($author3); $scoresum=$author3["sum"];}$sql3 = "select count(*) as count from ask_socre where qid='".$qid."' ";$result3=mysql_query($sql3);echo $sql3;if($author4 = mysql_fetch_array ( $result3, MYSQL_ASSOC )){ var_dump($author4); $scorenum=$author4["count"];}
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